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图</h1><div id="post-meta"><div class="meta-firstline"><span class="post-meta-date"><i class="far fa-calendar-alt fa-fw post-meta-icon"></i><span class="post-meta-label">发表于</span><time class="post-meta-date-created" datetime="2021-12-06T09:36:00.000Z" title="发表于 2021-12-06 17:36:00">2021-12-06</time><span class="post-meta-separator">|</span><i class="fas fa-history fa-fw post-meta-icon"></i><span class="post-meta-label">更新于</span><time class="post-meta-date-updated" datetime="2021-12-10T12:57:38.857Z" title="更新于 2021-12-10 20:57:38">2021-12-10</time></span><span class="post-meta-categories"><span class="post-meta-separator">|</span><i class="fas fa-inbox fa-fw post-meta-icon"></i><a class="post-meta-categories" href="/hexo3/categories/Leetcode%E9%A2%98%E8%A7%A3/">Leetcode题解</a></span></div><div class="meta-secondline"><span class="post-meta-separator">|</span><span class="post-meta-wordcount"><i class="far fa-file-word fa-fw post-meta-icon"></i><span class="post-meta-label">字数总计:</span><span class="word-count">1.1k</span><span class="post-meta-separator">|</span><i class="far fa-clock fa-fw post-meta-icon"></i><span class="post-meta-label">阅读时长:</span><span>5分钟</span></span><span class="post-meta-separator">|</span><span class="post-meta-pv-cv" id="" data-flag-title="Leetcode 题解 - 图"><i class="far fa-eye fa-fw post-meta-icon"></i><span class="post-meta-label">阅读量:</span><span id="busuanzi_value_page_pv"></span></span></div></div></div></header><main class="layout" id="content-inner"><div id="post"><article class="post-content" id="article-container"><h1 id="Leetcode-题解-图"><a href="#Leetcode-题解-图" class="headerlink" title="Leetcode 题解 - 图"></a>Leetcode 题解 - 图</h1><!-- GFM-TOC -->
<ul>
<li><a href="#leetcode-%E9%A2%98%E8%A7%A3---%E5%9B%BE">Leetcode 题解 - 图</a><ul>
<li><a href="#%E4%BA%8C%E5%88%86%E5%9B%BE">二分图</a><ul>
<li><a href="#1-%E5%88%A4%E6%96%AD%E6%98%AF%E5%90%A6%E4%B8%BA%E4%BA%8C%E5%88%86%E5%9B%BE">1. 判断是否为二分图</a></li>
</ul>
</li>
<li><a href="#%E6%8B%93%E6%89%91%E6%8E%92%E5%BA%8F">拓扑排序</a><ul>
<li><a href="#1-%E8%AF%BE%E7%A8%8B%E5%AE%89%E6%8E%92%E7%9A%84%E5%90%88%E6%B3%95%E6%80%A7">1. 课程安排的合法性</a></li>
<li><a href="#2-%E8%AF%BE%E7%A8%8B%E5%AE%89%E6%8E%92%E7%9A%84%E9%A1%BA%E5%BA%8F">2. 课程安排的顺序</a></li>
</ul>
</li>
<li><a href="#%E5%B9%B6%E6%9F%A5%E9%9B%86">并查集</a><ul>
<li><a href="#1-%E5%86%97%E4%BD%99%E8%BF%9E%E6%8E%A5">1. 冗余连接</a><!-- GFM-TOC --></li>
</ul>
</li>
</ul>
</li>
</ul>
<h2 id="二分图"><a href="#二分图" class="headerlink" title="二分图"></a>二分图</h2><p>如果可以用两种颜色对图中的节点进行着色，并且保证相邻的节点颜色不同，那么这个图就是二分图。</p>
<h3 id="1-判断是否为二分图"><a href="#1-判断是否为二分图" class="headerlink" title="1. 判断是否为二分图"></a>1. 判断是否为二分图</h3><p>785. Is Graph Bipartite? (Medium)</p>
<p><a target="_blank" rel="noopener" href="https://leetcode.com/problems/is-graph-bipartite/description/">Leetcode</a> / <a target="_blank" rel="noopener" href="https://leetcode-cn.com/problems/is-graph-bipartite/description/">力扣</a></p>
<figure class="highlight html"><table><tr><td class="code"><pre><span class="line">Input: [[1,3], [0,2], [1,3], [0,2]]</span><br><span class="line">Output: true</span><br><span class="line">Explanation:</span><br><span class="line">The graph looks like this:</span><br><span class="line">0----1</span><br><span class="line">|    |</span><br><span class="line">|    |</span><br><span class="line">3----2</span><br><span class="line">We can divide the vertices into two groups: &#123;0, 2&#125; and &#123;1, 3&#125;.</span><br></pre></td></tr></table></figure>

<figure class="highlight html"><table><tr><td class="code"><pre><span class="line">Example 2:</span><br><span class="line">Input: [[1,2,3], [0,2], [0,1,3], [0,2]]</span><br><span class="line">Output: false</span><br><span class="line">Explanation:</span><br><span class="line">The graph looks like this:</span><br><span class="line">0----1</span><br><span class="line">| \  |</span><br><span class="line">|  \ |</span><br><span class="line">3----2</span><br><span class="line">We cannot find a way to divide the set of nodes into two independent subsets.</span><br></pre></td></tr></table></figure>

<figure class="highlight java"><table><tr><td class="code"><pre><span class="line"><span class="function"><span class="keyword">public</span> <span class="keyword">boolean</span> <span class="title">isBipartite</span><span class="params">(<span class="keyword">int</span>[][] graph)</span> </span>&#123;</span><br><span class="line">    <span class="keyword">int</span>[] colors = <span class="keyword">new</span> <span class="keyword">int</span>[graph.length];</span><br><span class="line">    Arrays.fill(colors, -<span class="number">1</span>);</span><br><span class="line">    <span class="keyword">for</span> (<span class="keyword">int</span> i = <span class="number">0</span>; i &lt; graph.length; i++) &#123;  <span class="comment">// 处理图不是连通的情况</span></span><br><span class="line">        <span class="keyword">if</span> (colors[i] == -<span class="number">1</span> &amp;&amp; !isBipartite(i, <span class="number">0</span>, colors, graph)) &#123;</span><br><span class="line">            <span class="keyword">return</span> <span class="keyword">false</span>;</span><br><span class="line">        &#125;</span><br><span class="line">    &#125;</span><br><span class="line">    <span class="keyword">return</span> <span class="keyword">true</span>;</span><br><span class="line">&#125;</span><br><span class="line"></span><br><span class="line"><span class="function"><span class="keyword">private</span> <span class="keyword">boolean</span> <span class="title">isBipartite</span><span class="params">(<span class="keyword">int</span> curNode, <span class="keyword">int</span> curColor, <span class="keyword">int</span>[] colors, <span class="keyword">int</span>[][] graph)</span> </span>&#123;</span><br><span class="line">    <span class="keyword">if</span> (colors[curNode] != -<span class="number">1</span>) &#123;</span><br><span class="line">        <span class="keyword">return</span> colors[curNode] == curColor;</span><br><span class="line">    &#125;</span><br><span class="line">    colors[curNode] = curColor;</span><br><span class="line">    <span class="keyword">for</span> (<span class="keyword">int</span> nextNode : graph[curNode]) &#123;</span><br><span class="line">        <span class="keyword">if</span> (!isBipartite(nextNode, <span class="number">1</span> - curColor, colors, graph)) &#123;</span><br><span class="line">            <span class="keyword">return</span> <span class="keyword">false</span>;</span><br><span class="line">        &#125;</span><br><span class="line">    &#125;</span><br><span class="line">    <span class="keyword">return</span> <span class="keyword">true</span>;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure>

<h2 id="拓扑排序"><a href="#拓扑排序" class="headerlink" title="拓扑排序"></a>拓扑排序</h2><p>常用于在具有先序关系的任务规划中。</p>
<h3 id="1-课程安排的合法性"><a href="#1-课程安排的合法性" class="headerlink" title="1. 课程安排的合法性"></a>1. 课程安排的合法性</h3><p>207. Course Schedule (Medium)</p>
<p><a target="_blank" rel="noopener" href="https://leetcode.com/problems/course-schedule/description/">Leetcode</a> / <a target="_blank" rel="noopener" href="https://leetcode-cn.com/problems/course-schedule/description/">力扣</a></p>
<figure class="highlight html"><table><tr><td class="code"><pre><span class="line">2, [[1,0]]</span><br><span class="line">return true</span><br></pre></td></tr></table></figure>

<figure class="highlight html"><table><tr><td class="code"><pre><span class="line">2, [[1,0],[0,1]]</span><br><span class="line">return false</span><br></pre></td></tr></table></figure>

<p>题目描述：一个课程可能会先修课程，判断给定的先修课程规定是否合法。</p>
<p>本题不需要使用拓扑排序，只需要检测有向图是否存在环即可。</p>
<figure class="highlight java"><table><tr><td class="code"><pre><span class="line"><span class="function"><span class="keyword">public</span> <span class="keyword">boolean</span> <span class="title">canFinish</span><span class="params">(<span class="keyword">int</span> numCourses, <span class="keyword">int</span>[][] prerequisites)</span> </span>&#123;</span><br><span class="line">    List&lt;Integer&gt;[] graphic = <span class="keyword">new</span> List[numCourses];</span><br><span class="line">    <span class="keyword">for</span> (<span class="keyword">int</span> i = <span class="number">0</span>; i &lt; numCourses; i++) &#123;</span><br><span class="line">        graphic[i] = <span class="keyword">new</span> ArrayList&lt;&gt;();</span><br><span class="line">    &#125;</span><br><span class="line">    <span class="keyword">for</span> (<span class="keyword">int</span>[] pre : prerequisites) &#123;</span><br><span class="line">        graphic[pre[<span class="number">0</span>]].add(pre[<span class="number">1</span>]);</span><br><span class="line">    &#125;</span><br><span class="line">    <span class="keyword">boolean</span>[] globalMarked = <span class="keyword">new</span> <span class="keyword">boolean</span>[numCourses];</span><br><span class="line">    <span class="keyword">boolean</span>[] localMarked = <span class="keyword">new</span> <span class="keyword">boolean</span>[numCourses];</span><br><span class="line">    <span class="keyword">for</span> (<span class="keyword">int</span> i = <span class="number">0</span>; i &lt; numCourses; i++) &#123;</span><br><span class="line">        <span class="keyword">if</span> (hasCycle(globalMarked, localMarked, graphic, i)) &#123;</span><br><span class="line">            <span class="keyword">return</span> <span class="keyword">false</span>;</span><br><span class="line">        &#125;</span><br><span class="line">    &#125;</span><br><span class="line">    <span class="keyword">return</span> <span class="keyword">true</span>;</span><br><span class="line">&#125;</span><br><span class="line"></span><br><span class="line"><span class="function"><span class="keyword">private</span> <span class="keyword">boolean</span> <span class="title">hasCycle</span><span class="params">(<span class="keyword">boolean</span>[] globalMarked, <span class="keyword">boolean</span>[] localMarked,</span></span></span><br><span class="line"><span class="params"><span class="function">                         List&lt;Integer&gt;[] graphic, <span class="keyword">int</span> curNode)</span> </span>&#123;</span><br><span class="line"></span><br><span class="line">    <span class="keyword">if</span> (localMarked[curNode]) &#123;</span><br><span class="line">        <span class="keyword">return</span> <span class="keyword">true</span>;</span><br><span class="line">    &#125;</span><br><span class="line">    <span class="keyword">if</span> (globalMarked[curNode]) &#123;</span><br><span class="line">        <span class="keyword">return</span> <span class="keyword">false</span>;</span><br><span class="line">    &#125;</span><br><span class="line">    globalMarked[curNode] = <span class="keyword">true</span>;</span><br><span class="line">    localMarked[curNode] = <span class="keyword">true</span>;</span><br><span class="line">    <span class="keyword">for</span> (<span class="keyword">int</span> nextNode : graphic[curNode]) &#123;</span><br><span class="line">        <span class="keyword">if</span> (hasCycle(globalMarked, localMarked, graphic, nextNode)) &#123;</span><br><span class="line">            <span class="keyword">return</span> <span class="keyword">true</span>;</span><br><span class="line">        &#125;</span><br><span class="line">    &#125;</span><br><span class="line">    localMarked[curNode] = <span class="keyword">false</span>;</span><br><span class="line">    <span class="keyword">return</span> <span class="keyword">false</span>;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure>

<h3 id="2-课程安排的顺序"><a href="#2-课程安排的顺序" class="headerlink" title="2. 课程安排的顺序"></a>2. 课程安排的顺序</h3><p>210. Course Schedule II (Medium)</p>
<p><a target="_blank" rel="noopener" href="https://leetcode.com/problems/course-schedule-ii/description/">Leetcode</a> / <a target="_blank" rel="noopener" href="https://leetcode-cn.com/problems/course-schedule-ii/description/">力扣</a></p>
<figure class="highlight html"><table><tr><td class="code"><pre><span class="line">4, [[1,0],[2,0],[3,1],[3,2]]</span><br><span class="line">There are a total of 4 courses to take. To take course 3 you should have finished both courses 1 and 2. Both courses 1 and 2 should be taken after you finished course 0. So one correct course order is [0,1,2,3]. Another correct ordering is[0,2,1,3].</span><br></pre></td></tr></table></figure>

<p>使用 DFS 来实现拓扑排序，使用一个栈存储后序遍历结果，这个栈的逆序结果就是拓扑排序结果。</p>
<p>证明：对于任何先序关系：v-&gt;w，后序遍历结果可以保证 w 先进入栈中，因此栈的逆序结果中 v 会在 w 之前。</p>
<figure class="highlight java"><table><tr><td class="code"><pre><span class="line"><span class="keyword">public</span> <span class="keyword">int</span>[] findOrder(<span class="keyword">int</span> numCourses, <span class="keyword">int</span>[][] prerequisites) &#123;</span><br><span class="line">    List&lt;Integer&gt;[] graphic = <span class="keyword">new</span> List[numCourses];</span><br><span class="line">    <span class="keyword">for</span> (<span class="keyword">int</span> i = <span class="number">0</span>; i &lt; numCourses; i++) &#123;</span><br><span class="line">        graphic[i] = <span class="keyword">new</span> ArrayList&lt;&gt;();</span><br><span class="line">    &#125;</span><br><span class="line">    <span class="keyword">for</span> (<span class="keyword">int</span>[] pre : prerequisites) &#123;</span><br><span class="line">        graphic[pre[<span class="number">0</span>]].add(pre[<span class="number">1</span>]);</span><br><span class="line">    &#125;</span><br><span class="line">    Stack&lt;Integer&gt; postOrder = <span class="keyword">new</span> Stack&lt;&gt;();</span><br><span class="line">    <span class="keyword">boolean</span>[] globalMarked = <span class="keyword">new</span> <span class="keyword">boolean</span>[numCourses];</span><br><span class="line">    <span class="keyword">boolean</span>[] localMarked = <span class="keyword">new</span> <span class="keyword">boolean</span>[numCourses];</span><br><span class="line">    <span class="keyword">for</span> (<span class="keyword">int</span> i = <span class="number">0</span>; i &lt; numCourses; i++) &#123;</span><br><span class="line">        <span class="keyword">if</span> (hasCycle(globalMarked, localMarked, graphic, i, postOrder)) &#123;</span><br><span class="line">            <span class="keyword">return</span> <span class="keyword">new</span> <span class="keyword">int</span>[<span class="number">0</span>];</span><br><span class="line">        &#125;</span><br><span class="line">    &#125;</span><br><span class="line">    <span class="keyword">int</span>[] orders = <span class="keyword">new</span> <span class="keyword">int</span>[numCourses];</span><br><span class="line">    <span class="keyword">for</span> (<span class="keyword">int</span> i = numCourses - <span class="number">1</span>; i &gt;= <span class="number">0</span>; i--) &#123;</span><br><span class="line">        orders[i] = postOrder.pop();</span><br><span class="line">    &#125;</span><br><span class="line">    <span class="keyword">return</span> orders;</span><br><span class="line">&#125;</span><br><span class="line"></span><br><span class="line"><span class="function"><span class="keyword">private</span> <span class="keyword">boolean</span> <span class="title">hasCycle</span><span class="params">(<span class="keyword">boolean</span>[] globalMarked, <span class="keyword">boolean</span>[] localMarked, List&lt;Integer&gt;[] graphic,</span></span></span><br><span class="line"><span class="params"><span class="function">                         <span class="keyword">int</span> curNode, Stack&lt;Integer&gt; postOrder)</span> </span>&#123;</span><br><span class="line"></span><br><span class="line">    <span class="keyword">if</span> (localMarked[curNode]) &#123;</span><br><span class="line">        <span class="keyword">return</span> <span class="keyword">true</span>;</span><br><span class="line">    &#125;</span><br><span class="line">    <span class="keyword">if</span> (globalMarked[curNode]) &#123;</span><br><span class="line">        <span class="keyword">return</span> <span class="keyword">false</span>;</span><br><span class="line">    &#125;</span><br><span class="line">    globalMarked[curNode] = <span class="keyword">true</span>;</span><br><span class="line">    localMarked[curNode] = <span class="keyword">true</span>;</span><br><span class="line">    <span class="keyword">for</span> (<span class="keyword">int</span> nextNode : graphic[curNode]) &#123;</span><br><span class="line">        <span class="keyword">if</span> (hasCycle(globalMarked, localMarked, graphic, nextNode, postOrder)) &#123;</span><br><span class="line">            <span class="keyword">return</span> <span class="keyword">true</span>;</span><br><span class="line">        &#125;</span><br><span class="line">    &#125;</span><br><span class="line">    localMarked[curNode] = <span class="keyword">false</span>;</span><br><span class="line">    postOrder.push(curNode);</span><br><span class="line">    <span class="keyword">return</span> <span class="keyword">false</span>;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure>

<h2 id="并查集"><a href="#并查集" class="headerlink" title="并查集"></a>并查集</h2><p>并查集可以动态地连通两个点，并且可以非常快速地判断两个点是否连通。</p>
<h3 id="1-冗余连接"><a href="#1-冗余连接" class="headerlink" title="1. 冗余连接"></a>1. 冗余连接</h3><p>684. Redundant Connection (Medium)</p>
<p><a target="_blank" rel="noopener" href="https://leetcode.com/problems/redundant-connection/description/">Leetcode</a> / <a target="_blank" rel="noopener" href="https://leetcode-cn.com/problems/redundant-connection/description/">力扣</a></p>
<figure class="highlight html"><table><tr><td class="code"><pre><span class="line">Input: [[1,2], [1,3], [2,3]]</span><br><span class="line">Output: [2,3]</span><br><span class="line">Explanation: The given undirected graph will be like this:</span><br><span class="line">  1</span><br><span class="line"> / \</span><br><span class="line">2 - 3</span><br></pre></td></tr></table></figure>

<p>题目描述：有一系列的边连成的图，找出一条边，移除它之后该图能够成为一棵树。</p>
<figure class="highlight java"><table><tr><td class="code"><pre><span class="line"><span class="keyword">public</span> <span class="keyword">int</span>[] findRedundantConnection(<span class="keyword">int</span>[][] edges) &#123;</span><br><span class="line">    <span class="keyword">int</span> N = edges.length;</span><br><span class="line">    UF uf = <span class="keyword">new</span> UF(N);</span><br><span class="line">    <span class="keyword">for</span> (<span class="keyword">int</span>[] e : edges) &#123;</span><br><span class="line">        <span class="keyword">int</span> u = e[<span class="number">0</span>], v = e[<span class="number">1</span>];</span><br><span class="line">        <span class="keyword">if</span> (uf.connect(u, v)) &#123;</span><br><span class="line">            <span class="keyword">return</span> e;</span><br><span class="line">        &#125;</span><br><span class="line">        uf.union(u, v);</span><br><span class="line">    &#125;</span><br><span class="line">    <span class="keyword">return</span> <span class="keyword">new</span> <span class="keyword">int</span>[]&#123;-<span class="number">1</span>, -<span class="number">1</span>&#125;;</span><br><span class="line">&#125;</span><br><span class="line"></span><br><span class="line"><span class="keyword">private</span> <span class="class"><span class="keyword">class</span> <span class="title">UF</span> </span>&#123;</span><br><span class="line"></span><br><span class="line">    <span class="keyword">private</span> <span class="keyword">int</span>[] id;</span><br><span class="line"></span><br><span class="line">    UF(<span class="keyword">int</span> N) &#123;</span><br><span class="line">        id = <span class="keyword">new</span> <span class="keyword">int</span>[N + <span class="number">1</span>];</span><br><span class="line">        <span class="keyword">for</span> (<span class="keyword">int</span> i = <span class="number">0</span>; i &lt; id.length; i++) &#123;</span><br><span class="line">            id[i] = i;</span><br><span class="line">        &#125;</span><br><span class="line">    &#125;</span><br><span class="line"></span><br><span class="line">    <span class="function"><span class="keyword">void</span> <span class="title">union</span><span class="params">(<span class="keyword">int</span> u, <span class="keyword">int</span> v)</span> </span>&#123;</span><br><span class="line">        <span class="keyword">int</span> uID = find(u);</span><br><span class="line">        <span class="keyword">int</span> vID = find(v);</span><br><span class="line">        <span class="keyword">if</span> (uID == vID) &#123;</span><br><span class="line">            <span class="keyword">return</span>;</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">for</span> (<span class="keyword">int</span> i = <span class="number">0</span>; i &lt; id.length; i++) &#123;</span><br><span class="line">            <span class="keyword">if</span> (id[i] == uID) &#123;</span><br><span class="line">                id[i] = vID;</span><br><span class="line">            &#125;</span><br><span class="line">        &#125;</span><br><span class="line">    &#125;</span><br><span class="line"></span><br><span class="line">    <span class="function"><span class="keyword">int</span> <span class="title">find</span><span class="params">(<span class="keyword">int</span> p)</span> </span>&#123;</span><br><span class="line">        <span class="keyword">return</span> id[p];</span><br><span class="line">    &#125;</span><br><span class="line"></span><br><span class="line">    <span class="function"><span class="keyword">boolean</span> <span class="title">connect</span><span class="params">(<span class="keyword">int</span> u, <span class="keyword">int</span> v)</span> </span>&#123;</span><br><span class="line">        <span class="keyword">return</span> find(u) == find(v);</span><br><span class="line">    &#125;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure>
</article><div class="post-copyright"><div class="post-copyright__author"><span class="post-copyright-meta">文章作者: </span><span class="post-copyright-info"><a href="mailto:undefined">Zhang Shuo</a></span></div><div class="post-copyright__type"><span class="post-copyright-meta">文章链接: </span><span class="post-copyright-info"><a href="https://zhang-shuo-fr.gitee.io/hexo3/2021/12/06/notes/Leetcode%20%E9%A2%98%E8%A7%A3%20-%20%E5%9B%BE/">https://zhang-shuo-fr.gitee.io/hexo3/2021/12/06/notes/Leetcode%20%E9%A2%98%E8%A7%A3%20-%20%E5%9B%BE/</a></span></div><div class="post-copyright__notice"><span class="post-copyright-meta">版权声明: </span><span class="post-copyright-info">本博客所有文章除特别声明外，均采用 <a href="https://creativecommons.org/licenses/by-nc-sa/4.0/" target="_blank">CC BY-NC-SA 4.0</a> 许可协议。转载请注明来自 <a href="https://zhang-shuo-fr.gitee.io/hexo3" target="_blank">Zhang Shuo'blog</a>！</span></div></div><div class="tag_share"><div class="post-meta__tag-list"><a class="post-meta__tags" href="/hexo3/tags/Leetcode%E9%A2%98%E8%A7%A3/">Leetcode题解</a></div><div class="post_share"><div class="social-share" data-image="/hexo3/img/19.jpg" data-sites="facebook,twitter,wechat,weibo,qq"></div><link rel="stylesheet" href="https://cdn.jsdelivr.net/npm/social-share.js/dist/css/share.min.css" media="print" onload="this.media='all'"><script src="https://cdn.jsdelivr.net/npm/social-share.js/dist/js/social-share.min.js" defer></script></div></div><div class="post-reward"><div class="reward-button button--animated"><i class="fas fa-qrcode"></i> 打赏</div><div class="reward-main"><ul class="reward-all"><li class="reward-item"><a href="/hexo3/img/wechat.jpg" target="_blank"><img class="post-qr-code-img" src= "" data-lazy-src="/hexo3/img/wechat.jpg" alt="微信"/></a><div class="post-qr-code-desc">微信</div></li><li class="reward-item"><a href="/hexo3/img/alipay.jpg" target="_blank"><img class="post-qr-code-img" src= "" data-lazy-src="/hexo3/img/alipay.jpg" alt="支付宝"/></a><div class="post-qr-code-desc">支付宝</div></li></ul></div></div><nav class="pagination-post" id="pagination"><div class="prev-post pull-left"><a href="/hexo3/2021/12/06/notes/Leetcode%20%E9%A2%98%E8%A7%A3%20-%20%E6%8E%92%E5%BA%8F/"><img class="prev-cover" src= "" data-lazy-src="/hexo3/img/7.jpg" onerror="onerror=null;src='/hexo3/img/404.jpg'" alt="cover of previous post"><div class="pagination-info"><div class="label">上一篇</div><div class="prev_info">Leetcode 题解 - 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位运算"><img class="cover" src= "" data-lazy-src="/hexo3/img/18.jpg" alt="cover"><div class="content is-center"><div class="date"><i class="far fa-calendar-alt fa-fw"></i> 2021-12-06</div><div class="title">Leetcode 题解 - 位运算</div></div></a></div><div><a href="/hexo3/2021/12/06/notes/Leetcode%20%E9%A2%98%E8%A7%A3%20-%20%E5%88%86%E6%B2%BB/" title="Leetcode 题解 - 分治"><img class="cover" src= "" data-lazy-src="/hexo3/img/8.jpg" alt="cover"><div class="content is-center"><div class="date"><i class="far fa-calendar-alt fa-fw"></i> 2021-12-06</div><div class="title">Leetcode 题解 - 分治</div></div></a></div><div><a href="/hexo3/2021/12/06/notes/Leetcode%20%E9%A2%98%E8%A7%A3%20-%20%E5%8F%8C%E6%8C%87%E9%92%88/" title="Leetcode 题解 - 双指针"><img class="cover" src= "" data-lazy-src="/hexo3/img/1.jpg" alt="cover"><div class="content is-center"><div class="date"><i class="far fa-calendar-alt fa-fw"></i> 2021-12-06</div><div class="title">Leetcode 题解 - 双指针</div></div></a></div><div><a href="/hexo3/2021/12/06/notes/Leetcode%20%E9%A2%98%E8%A7%A3%20-%20%E5%93%88%E5%B8%8C%E8%A1%A8/" title="Leetcode 题解 - 哈希表"><img class="cover" src= "" data-lazy-src="/hexo3/img/12.jpg" alt="cover"><div class="content is-center"><div class="date"><i class="far fa-calendar-alt fa-fw"></i> 2021-12-06</div><div class="title">Leetcode 题解 - 哈希表</div></div></a></div><div><a href="/hexo3/2021/12/06/notes/Leetcode%20%E9%A2%98%E8%A7%A3%20-%20%E6%8E%92%E5%BA%8F/" title="Leetcode 题解 - 排序"><img class="cover" src= "" data-lazy-src="/hexo3/img/7.jpg" alt="cover"><div class="content is-center"><div class="date"><i class="far fa-calendar-alt fa-fw"></i> 2021-12-06</div><div class="title">Leetcode 题解 - 排序</div></div></a></div></div></div></div><div class="aside-content" id="aside-content"><div class="sticky_layout"><div class="card-widget" id="card-toc"><div class="item-headline"><i class="fas fa-stream"></i><span>目录</span></div><div class="toc-content"><ol class="toc"><li class="toc-item toc-level-1"><a class="toc-link" href="#Leetcode-%E9%A2%98%E8%A7%A3-%E5%9B%BE"><span class="toc-number">1.</span> <span class="toc-text">Leetcode 题解 - 图</span></a><ol class="toc-child"><li class="toc-item toc-level-2"><a class="toc-link" href="#%E4%BA%8C%E5%88%86%E5%9B%BE"><span class="toc-number">1.1.</span> <span class="toc-text">二分图</span></a><ol class="toc-child"><li class="toc-item toc-level-3"><a class="toc-link" href="#1-%E5%88%A4%E6%96%AD%E6%98%AF%E5%90%A6%E4%B8%BA%E4%BA%8C%E5%88%86%E5%9B%BE"><span class="toc-number">1.1.1.</span> <span class="toc-text">1. 判断是否为二分图</span></a></li></ol></li><li class="toc-item toc-level-2"><a class="toc-link" href="#%E6%8B%93%E6%89%91%E6%8E%92%E5%BA%8F"><span class="toc-number">1.2.</span> <span class="toc-text">拓扑排序</span></a><ol class="toc-child"><li class="toc-item toc-level-3"><a class="toc-link" href="#1-%E8%AF%BE%E7%A8%8B%E5%AE%89%E6%8E%92%E7%9A%84%E5%90%88%E6%B3%95%E6%80%A7"><span class="toc-number">1.2.1.</span> <span class="toc-text">1. 课程安排的合法性</span></a></li><li class="toc-item toc-level-3"><a class="toc-link" href="#2-%E8%AF%BE%E7%A8%8B%E5%AE%89%E6%8E%92%E7%9A%84%E9%A1%BA%E5%BA%8F"><span class="toc-number">1.2.2.</span> <span class="toc-text">2. 课程安排的顺序</span></a></li></ol></li><li class="toc-item toc-level-2"><a class="toc-link" href="#%E5%B9%B6%E6%9F%A5%E9%9B%86"><span class="toc-number">1.3.</span> <span class="toc-text">并查集</span></a><ol class="toc-child"><li class="toc-item toc-level-3"><a class="toc-link" href="#1-%E5%86%97%E4%BD%99%E8%BF%9E%E6%8E%A5"><span class="toc-number">1.3.1.</span> <span class="toc-text">1. 冗余连接</span></a></li></ol></li></ol></li></ol></div></div></div></div></main><footer id="footer" style="background-image: url('/hexo3/img/19.jpg')"><div id="footer-wrap"><div class="copyright">&copy;2020 - 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